The Logic behind Array.prototype.includes()

JavaScript
1

Tell us what’s happening:
Hi everyone,
Even though this code worked as wanted, I couldn’t understand the logic under the Array.prototype.includes() method.
I indicated the array the one I want to work on it with the name of ‘list’. However, I did not specify the variable in the Array.prototype.includes() method, which is ‘val’. So, how did the JS understood the value of the variable as the ones comes after the arguments[0].

I’m a newbie learner, so do not hesitate to correct me in terms of terminology that I used, and thanks for your answer.

Your code so far


function destroyer(arr) {
let list = arguments[0];
return list.filter(val => !Object.values(arguments).slice(1).includes(val));

}

destroyer([1, 2, 3, 1, 2, 3], 2, 3);

Your browser information:

User Agent is: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/84.0.4147.105 Safari/537.36.

Challenge: Seek and Destroy

Link to the challenge:

2

arguments[0] is already named variable arr and you can run following to reassure yourself:

console.log(arr === list); // true

Then this:

can simply be substituted by following and hopefully it will make a bit more sense in understanding the role of includes() here:

![...arguments]